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2-1ECE ET LABMANUAL
Dept of E.E.E Page 1
SERIES AND PAREALLEL RESONANCE
AIM: To determine the performance of the series and parallel circuit at resonance.
SERIES RESONANCE
GIVEN CIRCUIT: MODEL GRAPH:
PARALLEL RESONANCE
Exp - 1
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 2
CIRCUIT DIAGRAMS:
FOR SERIES RESONANCE :
FOR PARALLEL RESONANCE:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 3
APPARATUS:-
S. No Name of the apparatus Range Type Quantity
1 Signal generator (0 – 1M)Hz,
(0-20) VPP digital 01No
2 Decade inductance Box (0-40mH) - 01No
3 Voltmeters (0-30V) MI 03No
4 Capacitor 0.1uf - 01No
5 Resistors 30 Ω - 01No
6 Ammeter (0-200m) A MI 01No
7 Experimental board - - 1No
8 C.R.O 30 MHz - 1No
9 Connecting wires - - Required
Number
THEORY:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 4
Resonance is a particular type of phenomenon inherently found normally in every
kind of system, electrical, mechanical, optical, Acoustical and even atomic. There are several
definitions of resonance. But, the most frequently used definition of resonance in electrical
system is studied state operation of a circuit or system at that frequency for which the resultant
response is in time phase with the forcing function.
Series resonance:
A circuit is said to be under resonance, when the applied voltage ‘V’ and current are in
phase. Thus a series RLC circuit, under resonance behaves like a pure resistance network and the
reactance of the circuit should be zero. Since V & I are in phase, the power factor is unity at
resonance.
The frequency at which the resonance will occur is known as resonant frequency.
Resonant frequency, fr =
√
Thus at resonance the impedance Z is minimum. Since I = V/Z. The current is maximum . So
that current amplification takes place.
Parallel Resonance:
The parallel circuit consisting branches with single pure elements R,L & C is
an ideal circuit. How ever the performance of such a circuit is of interest in the general subject
of resonance. This ideal parallel circuit is of interest in the general subject of resonance.
Lower cut-off frequency is above the resonant frequency at which the current is reduced
to
√ times of it’s minimum value. Upper cut-off frequency is above.
Quality factor is the ratio of reactance power inductor (or) capacitor to its resistance. Selectivity
is the reciprocal of the quality factors.
PROCEDURE:
1. Connections are made as per the circuit diagram.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 5
2. By varying the frequency note down the corresponding values of current in both cases and
note down f, VC, VL and VR.
3. At a particular value of frequency the current reaches its Maximum /minimum in
Series/Parallel resonance. That instant of frequency VC = VL and VR = VS in series Resonance
circuit.
THEORITICAL CALCULATIONS:
For Series Resonance circuit:
1. Resonant frequency fr =
√
=2516 Hz
2. Lower cut-off frequency fl = +
= 2457.85 Hz
3. Upper cut-off frequency f2 =
+
= 2576.14 Hz
4. Band width = f2-f1 = 119.66 Hz
5. Quality factor Q =
= .
= 21.07
6. Current at Resonance Io = VRo/R
TABULAR FORMS:
FOR SERIES RESONANCE :
S.NO FREQUNCY V(L) V(C) V(R) I(mA)
1
2
...
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 6
TABULAR FORMS:
FOR PARALLEL RESONANCE:
S.NO FREQUNCY V(L) V(C) V(R) I(mA)
1
2
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 7
For parallel Resonance circuit:
1. Resonant frequency fr =
= 2514 Hz
2. Lower cut-off frequency fl =
= 2456. 77 Hz
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 8
3. Upper cut-off frequency f2 =
=2577.6 Hz
4. Band width =
= 120.07 Hz 5. Quality factor Q =
= 0.0474 Hz
6. Current at resonance Io = VRo/R
PRECAUTIONS:
• Meter reading should be taken with out parallax error.
• Connection should be made tight.
RESULT:-
Review Questions:- 1. Definition of resonance?
2. Define the series resonance?
3. Define the parallel resonance?
4. Applications of resonance?
5. What is the condition of voltage ¤t at the resonance condition?
CONCLUSIONS: 1. Since the current at resonance is maximum, the series resonant circuit is
called as acceptor circuit.
2. As the resistance of the circuit decreases, the Q-factor increases and
selectivity of the circuit will be better.
3. Since the current at resonance is minimum, the parallel resonant circuit is
called as rejector circuit.
4. The variation of the resistance does not affect the resonant frequency.
TWO PORT NETWORK PARAMETERS
AIM:
S.No Parameter
Series Resonant circuit Parallel Resonant circuit
Theoretical
Values
Practical
Values
Theoretical
Values
Practical
Values
1 Resonant
Frequency, fr 2516 Hz 2514 Hz
2 Band width 119.66 Hz 120.07 Hz
3 Quality factor 21.07 0.0474 Hz
Exp - 2
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 9
To determine open circuit impedance parameters (Z) and short circuit admittance
parameters (Y) of the given two port network .
.
GIVEN CIRCUIT:
APPARATUS:
S. No Name of the apparatus Range Type Quantity
1 Regulated power supply (0 – 30) V/2A digital 01
2 Voltmeters (0-30) V MC 01
3 Ammeters (0-200m)A MC 01
4 Resistors
330 Ω
470Ω
630Ω
-
01
01
01
5 Experimental board - - 01
6 Connecting wires - - Required
Number
CIRCUIT DIAGRAMS:
BASIC CIRCUIT
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 10
WHEN V1=0
WHEN I1=0
WHEN V2=0
WHEN I2=0
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 11
THEORY:
A port is normally referred to a pair of terminals of a network though which we
can have access to network of calculating current in any part of network. Frequently the problem
is move restried in nature and may be that of calculating the response at a terminal pair
designated an input when excitation is applied at another terminal pair designated as input
terminals. It is a problem of terminal through which it is accessible, is called “Two Port
Network.“
If we relate the voltage of one port to the current of the same port, we get driving
point immitance. On the other hand, if we relate the voltage of one port to the current at another
port, we get transfer immittance. Immitance is a general term used to represent either the
impedance or the admittance of a network.
We will consider a general two-port network composed of linear, bilateral
elements and no independent sources. Dependent sources are permitted. It is represented as a
black box with two accessible terminals pairs as shown in. The voltage and current at port -1 are
V1 and I1 and at port =II are V2 and I2. The position of V1 and V2 and the directions of I1 and I2
are customarily selected. out of four variables, I1I1V2 and I2 only two are independent. The other
two are expressed in terms of the independent variable of network parameters.
PROCEDURE:
1. Connections are made as per the circuit diagram.
2. Open the port – I i.e, I1=0 find the values of I1,I2, V1.
3. Short circuits the port V2 =0 find the values of V2,I1, I2.
4. Repeat steps 2,3 for port – II and find the values of V1,I1,I2 and V2,I1,I2 respectively.
5. Find all the parameters of two port networks I,e, Z,Y, ABCD, AI B
I C
I D
I, h, g
parameters from the above data.
PRECAUTIONS:
1. Initially keep the RPS output voltage knob in zero volt position.
2. Set the ammeter pointer to zero position.
3. Take the readings without parallax error.
4. Avoid loose connections.
5. Do not short-circuit the RPS output terminals.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 12
TABULAR FORMS:
Theoritical Values
V1
(volts)
I1
(mA)
V2
(mA)
I2
(mA)
V1=0
0 61 20 92
I1=0
11.42 0 20 57
V2=0
20 107 0 61
I2=0
20 66 13.2 0
Practical Values
V1
(volts)
I1
(mA)
V2
(mA)
I2
(mA)
V1=0
I1=0
V2=0
I2=0
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 13
Theoretical calculations:
1. When I1 = 0 (i.e.,) When port is open circuited:
RL = 216.66 ohms
V2 = 20 V
I2= = 57 mA
V1 = I2.R = 11.42 V
2. When port -2 is open circuited (I2=0):
V1 = 20V
Rt = 185.74 ohms
I1 = = 66 mA
V2 = 13.2V
3. When port -1 is short circuited ( V1=0):
V2 = 20 V
Rt = 216.66 ohms
I2 = = 92 mA
I1 = I2 ! "# - 61 mA
4. When port –II is short – circuited (V2 = 0 ) :
V1 = 20 V
Rt = 185.74 ohms
I1 = V1/Rt = 107 mA
I2 = 61 mA
calculations for parameters:
Z-parameters:
Z11 =$ / I2=0 =
Z12 =$ / I1=0 =
Z21 =$ / I2=0 =
Z22 =$ / I1=0 =
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 14
Y – Parameters
Y11 = $ / V2=0 =
Y12 = $ / V1=0 =
Y21 = $ / V2= 0 =
Y22 = $ / V1= 0 =
ABCD parameters:
A = / I2= 0 =
B =
$ / V2= 0 =
C = $ / I2= 0 =
D = $$ / V1=0 =
H – Parameters:
h11 = $ / V2 =0 =
h12 = / I1=0 =
h21 = $$ / V2 =0 =
h22 = $ / I1 =0 =
g- parameters:
g11 = $ / I2 =0 =
g12 = $$ / V1=0 =
g21 = / I2=0 =
g22 = $ / V1=0 =
A1B
1C
1D
1 parameters:
A1 =
/ I1=0 =
B1 =
$ / V1=0 =
C1 =-
$ / I1=0 =
D1 =
$$ / V1=0 =
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 15
RESULT:
S. No Parameter Theoretical Values Practical Values
1 Z11 303. Ω
2 Z12 199.99 Ω
3 Z21 200 Ω
4 Z22 350 Ω
5 Y11 5.38 m mhos
6 Y12 3.07 m mhos
7 Y21 3.05 m mhos
8 Y22 4.6 m mhos
Reviw Questions:-
1. Write the 2-port network equations in terms of hybrid parameter?
2. Define image impedance?
3. What is Z- parameter?
4. Write the network equations of Y- parameter
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 16
CONCLUSIONS:
1. Since Z12 = Z21 and Y12 = Y21 the given circuit is reciprocal.
2. Since Z11 = Z22 and Y11 = Y22 the given circuit is symmetrical.
3. There is a small deviation between theoretical and practical values because internal
resistances of source and meters are not considered.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 17
SUPERPOSITION AND RECIPROCITY THEOREM
Exp - 3
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 18
3(a).VERIFICATION OF SUPERPOSITION THEOREM
AIM: To verify the superposition theorem.
GIVEN CIRCUIT:
STATEMENT:
SuperPosition Theorem:
In any linear, bilateral, multi source network the response in any
element is equal to the algebraic sum of the responses obtained by each source acting
separately while all other sources are set equal to zero.
APPARATUS:
S. No Name of the apparatus Range Type Quantity
1 Dual channel regulated
power supply
(0 – 30)
V/2A digital 01No
2 Ammeter (0 – 200m) A MC 01No
3 Resistors
100Ω
150Ω
200Ω
-
01No
01No
01No
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 19
4 Experimental board - - 01No
5 Connecting wires - - Required number
CIRCUIT DIAGRAM:
WhenV1&V2 source acting(To find I):-
WhenV1 source acting(To find I1):-
WhenV2 source acting(To find I2):-
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 20
THEORY:-
Superposition theorem:-
Superposition theorem states that in a linear bilateral network consisting N number of
sources each branch current is the algebraic sum of N currents through ( branch voltage), each of
which is determined by considering one source at a time and removing all other sources. In
removing the sources, voltage sources are short circuited or replaced by resistances equal to their
internal resistances for no ideal sources, while the ideal current sources are open circuited.
PROCEDURE:
1. Connect the circuit as per the fig (1).
2. Adjust the output voltage of sources X and Y to appropriate values (Say 30V and20V
respectively).
3. Note down the response (current, IL) through the branch of interest i.e. AB ammeter
reading.
4. Now set the source Y (20V) to 0V.
5. Note down the response (current, ILl) through the branch AB (ammeter reading).
6. Now set the source X (20V) to 0V and source Y to 20V.
7. Note down the response (current, ILll) through the branch AB (ammeter reading).
8. Reduce the output voltage of the sources X and Y to 0V and switch off the supply.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 21
9. Disconnect the circuit.
TABULAR FORMS:
From Fig(1)
S. No
Applied
voltage
(V1) Volt
Applied
voltage
(V2) Volt
Current
IL
(mA)
From Fig(2)
S. No Applied voltage
(V1) Volt
Current
IL
(mA)
From Fig(3)
S. No Applied voltage
(V2) Volt
Current
IL
(mA)
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 22
THEORITICAL CALCULATIONS
From Fig(2)
I1=V1/(R1+(R2//R3)) = 161 mA
ILl =I*R2/(R2+R3) = 69 mA
From Fig(3)
I2=V2/(R2+(R1//R3)) = 92 mA
ILl1
=I*R1/(R1+R3) = 30 mA ; IL = ILl
+ ILl1
=69 mA+30 mA =99mA
PRECAUTIONS:
1. Initially keep the RPS output voltage knob in zero volt position.
2. Set the ammeter pointer at zero position.
3. Take the readings without parallax error.
4. Avoid loose connections.
5.Avoid short circuit of RPS output terminals.
RESULT:
S.No Load current Theoretical Values Practical Values
1 When Both sources are acting, IL 99 mA
2 When only source X is acting, ILl 69 mA
3 When only source Y is acting, ILll 11
LI 30 mA
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 23
Review Questions:-
1) What do you man by Unilateral and Bilateral network? Give the limitations of
Superposition theorem.
2) What are the equivalent internal impedances for an ideal voltage source and for a
Current source?
3) Transform a physical voltage source into its equivalent current source.
4) If all the 3 star connected impedance are identical and equal to ZA, then what is the
Delta connected resistors?
CONCLUSION:
1. The given circuit is linear, since the response is algebraic sum of the individual
responses.
2. Superposition theorem is not valid for power responses.
3( b) RECIPROCITY THEOREM
AIM :
To verify reciprocity theorem for the given circuit .
GIVEN CIRCUIT:
STATEMENT:
Reciprocity theorem
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 24
In any linear, bilateral, single source network, the ratio of excitation to the response is same
even though the positions of excitation and response are interchanged.
APPARATUS:
S. No Name of the apparatus Range Type Quantity
1 Regulated power supply (0 – 30) V/2A digital 01No
2 Ammeter (0 – 200m) A MC 01No
3 Resistors
100Ω
150Ω
200Ω
-
01 No
01No
01No
4 Experimental board - - 01No
5 Connecting wires - - Required number
CIRCUIT DIAGRAMS:-
CIRCUIT – 1:
CIRCUIT -2
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 25
THEORY:-
Reciprocity Theorems:-
This theorem permits in to transfer source from one position in the circuit to another and
may be stated as under.
In any linear bilateral network, if an e.m.f E acting in a branch causes a current ‘I’ in
branch ‘Y’ then the same e.m.f E located in branch ‘Y’ will cause a current I in branch.
However, currents in other branches will not change.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 26
PROCEDURE:-
Reciprocity Theorems:-
1. Connect the circuit as per the fig (1).
2. Adjust the output voltage of the regulated power supply to an appropriate value (Say
30V).
3. Note down the response (current, IL) through 150Ω resistor (ammeter reading) .
4. Reduce the output voltage of the signal generator to 0V and switch-off the supply.
5. Disconnect the circuit and connect the circuit as per the fig (2).
6. Adjust the output voltage of the regulated power supply to an appropriate value (Say
30V).
7. Note down the response (current, IL1) through 100Ω resistor (ammeter reading) .
8. Reduce the output voltage of the signal generator to 0V and switch-off the supply.
9. Disconnect the circuit.
TABULAR FORM:
RECIPROCITY THEOREM :
From Fig(1)
S. No Applied voltage
(V1) Volt
Current
IL
(mA)
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 27
From Fig(2)
S. No Applied voltage
(V2) Volt
Current
IL1
(mA)
THEORITICAL CALCULATIONS :
From Fig(1)
I1=V/(R1+(R2//R3)) = 161 mA
IL= I1*R3/(R2+R3) = 92 mA
From Fig(2)
I2=V/(R2+(R1//R3)) = 138 mA
IL1= I2*R3/(R1+R3) = 92 mA
PRECAUTIONS:
1. Initially keep the RPS output voltage knob in zero volt position.
2. Set the ammeter pointer at zero position.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 28
3. Take the readings without parallax error.
4. Avoid loose connections.
5.Avoid short circuit of RPS output terminals.
6.If voltmeter gives (-) ve reading then interchange the terminals connections of a voltmeter.
RESULT:
CONCLUSION:
1. The given circuit is bilateral, since the ratio of excitation to the response is same before
and after interchanging the positions of excitation and response.
S.No Parameter
Theoretical Value Practical Value
1 V/IL 92 mA
2 V/ILl 92 mA
Exp - 4
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 29
VERIFICATION OF MAXIMUM POWER TRANSFER THEOREM
AIM: To verify maximum power transfer theorem on d.c circuit .
STATEMENT:
It states that, maximum power will be transferred from source to load when the load
resistance is the complex conjugate of source resistance.
GIVEN CIRCUIT:
APPARATUS:
S. No Name of the apparatus Range Type Quantity
1 Regulated power supply (0 – 30)V/2A digital 01No
2 Voltmeter
(0-30) V MC 01No
3 Ammeter (0-1) A MC 01No
4 Rheostats 100 Ω/5A
50Ω/5A Wound Wire
01No
01No
5 Connecting wires - - Required number
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 30
CIRCUIT DIAGRAM:-
Fig(1)
Model Graph for Maximum Power Transfer Theorem:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 31
THEORY:
The statement of maximum power transfer is “ In d.c circuits, maximum power is
transferred from a source to load when the load resistance is made equal to the internal resistance
of the source as viewed from the load terminal with load removed and all e.m.f sources replaced
by their internal resistance.
Consider a voltage source of V of internal resistance R delivering power to aload
RL. We shall prove that when RL = RS the power transferred is maximum.
Circuit current = %
&'"&(
Power delivered P = I2 RL
= ) %&'"&*+
RL
,&' &(- ,&' &( &'- =0
RL+Ri cannot be zero,
Ri – RL = 0
PROCEDURE:
1. Connect the circuit as per the circuit diagram fig(1).
2. Adjust the output voltage of the regulated power supply to an appropriate value (Say
30V).
3. Vary the load rheostat. in steps, and note down the response (current) through the load for
each step (ammeter reading) & load voltage.
4. Reduce the output voltage of the regulated power supply to 0V and switch-off the supply.
5. Disconnect the circuit.
6. Calculate the power absorbed by the load, PL for each step using the formula PL=IL2
RL.
7. Plot the graph by taking ‘RL’ on X-axis and PL on Y-axis.
RS ==RL
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 32
8. Get the practical value of the load resistance for which it will gain the maximum power
from the source.
Tabulation for Maximum power transfer theorem:
SNO.
LOAD
RESISTENCE
In ohms
VOLTAGE
VL (in Volts)
CURRENT
IL (in amps)
POWER
P=VL*IL(watt)
1
2
…….
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 33
Theoretical calculations
Total current I = ./
&*"&'
P = I2 RL
= 0*&*"&'2. RL
Power is maximum dp/dRL =0
→ 1
1&' 2 ./&/"&' . &'3 = 0
→0/ 2 0/&/"&' &' 4&/ &'53 =0
→4&/ &'5 &/&' &' =0
&/ &' &' =
&/ &' =
&/ = &'
PRECAUTIONS:
→ Avoid loose connections
→ Ammeter should always connected in series with the circuit.
Rs=RL
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 34
RESULT:-
Review Questions:-
1. Derive the condition for maximum power transfer theorem.
2. Where and why maximum power transfer theorem is applied?
3. What is the efficiency of the circuit at the maximum power transfer
Condition & why?
4. Derive the condition for maximum power transfer theorem for a.c.
Circuits.
5.Define a dependent source.
VERIFICATION OF THEVENIN’S THEOREM AND NORTON’S THEOREM
AIM: To verify Thevenin’s & Norton’s theorems for the given circuit.
GIVEN CIRCUIT:
STATEMENTS:
Thevenin’s theorem
It states that any linear, active network with two open terminals can be
replaced by an equivalent circuit consisting of Thevenin’s equivalent voltage source Vth in series
with Thevenin’s equivalent resistance Rth. Where Vth is the open circuit voltage across the two
terminals and Rth is the resistance seen from the same two terminals.
Exp - 5
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 35
Norton’s theorem
It states that any linear, active network with two open terminals can be
replaced by an equivalent circuit consisting of Norton’s equivalent current source IN in parallel
with Norton’s equivalent resistance RN. where IN is the short circuit current through the two
terminals and RN is the resistance seen from the same two terminals
APPARATUS:
S. No Name of the
apparatus Range Type Quantity
1 Regulated power
supply
(0 –
30)V/2A Digital 01
2 Voltmeter
(0-30)V MC 01
3 Ammeter
(0-
2000m)A MC 01
4 Resistors
100Ω
150Ω
200Ω
Carbon
Composition
02
01
01
5 Experimental board --- --- 01
6 Connecting wires --- ---- Required
number
CIRCUIT DIAGRAMS:-
TO FIND IL:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 36
FIG(1)
TO FIND VTH:
FIG(2)
TO FIND Rth:
TO FIND IN:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 37
Fig(4)
THEORY:
Thevenin’s theorem:
The values of VTh and RTh are determined as mentioned in thevenin’s theorem.
Once the thevenin equivalent circuit is obtained, then current through any load resistance RL
connected across AB is given by, I = %6
&6"&'
Thevenin’s theorem is applied to d.c. circuits as stated below.
Any network having terminals A and B can be replaced by a single source of
e.m.f. VTh in series with a source resistance RRh.
(i) The e.m.f the voltage obtained across the terminals A and B with load, if any
removed i.e., it is open circuited voltage between terminals A and B.
(ii) The resistance RTh is the resistance of the network measured between the terminals A
and B with load removed and sources of e.m.f replaced by their internal resistances.
Ideal voltage sources are replaced with short circuits and ideal current sources are
replaced with open circuits.
To find VTh, the load resistor ‘RL’ is disconnected, then VTh = %
&"&7 Χ R3
To find RTh,
RTh = R2 + & &7&" &7
Thevenin’s theorem is also called as “Helmoltz theorem”
NORTON’S THEOREM:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 38
Nortom’s theorem is applied to d.c circuits may be stated as below.
Any linear network having two terminals ‘A’ and ‘B’ can be replaced by a current source
of current output IN in parallel with a resistance RN.
(i) The output IN of the current source is equal to the current that would flow through
AB when A&B are short circuited.
(ii) The resistance RN is the resistance of network measured b/wn A and B with load
removed and the sources of e.m.f replaced by their internal resistances.
Ideal voltage source are replaced with short circuits and ideal current sources are replaced
with open circuits .
Norton’s theorem is converse of thevenin’s theorem in that Norton equivalent
circuit uses a current generator instead of voltage generator and the resistance RN is parallel
with generator instead of being series with it.
for source current,
II = %&8 =
%4& "&7 5&&"&&7"&&7
for short-circuit current,
IN = Χ &7
&"&7 = %&7
&&"&&7"&&7
PROCEDURE:
Thevenin’s Theorem
1. Connect the circuit as per fig (1)
2. Adjust the output voltage of the regulated power supply to an appropriate value (Say
30V).
3. Note down the response (current, IL) through the branch of interest i.e. AB (ammeter
reading).
4. Reduce the output voltage of the regulated power supply to 0V and switch-off the supply.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 39
5. Disconnect the circuit and connect as per the fig (2).
6. Adjust the output voltage of the regulated power supply to 30V.
7. Note down the voltage across the load terminals AB (Voltmeter reading) that gives Vth.
8. Reduce the output voltage of the regulated power supply to 0V and switch-off the supply.
9. Disconnect the circuit and connect as per the fig (3).
10. Adjust the output voltage of the regulated power supply to an appropriate value (Say V =
30V).
11. Note down the current (I) supplied by the source (ammeter reading).
12. The ratio of V and I gives the Rth.
13. Reduce the output voltage of the regulated power supply to 0V and switch-off the supply.
14. Disconnect the circuit and connect as per the fig (4).
15. Adjust the output voltage of the regulated power supply to 30V
16. Note down the response (current, IN) through the branch AB (ammeter reading).
17. Reduce the output voltage of the regulated power supply to 0V and switch-off the supply.
18. Disconnect the circuit.
THEORITICAL CALCULATIONS :
Thevinen’s theorem:
VTH=(V/(R1+R3) )R3 = 20 V
RTH =((R1*R3)/(R1+R3))+RL = 216.6 Ω
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 40
IL= VTH / (RTH +RL) = 63 mA
Norton’s theorem:
RN = R2 +(R1*R3)/(R1+R3) = 216.6 Ω
VTH = V /(R1+R3) ) R3 = 20 V
IN= VTH /RTH = 92 mA
IL = IN *RTH/(RTH+RL) = 63 mA
Tabulation for Thevinen’s theorem:
THEORITICAL VALUES PRACTICAL VALUES
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 41
RTH=RL= 216.6 Ω
IL=63 mA
Vth=20 V
RTH=RL=
IL=
Vth=
Tabulation for Norton’s theorem:
THEORITICAL VALUES PRACTICAL VALUES
RN=RL=216.6 Ω
IL=63 mA
IN=92 mA
RN=RL=
IL=
IN=
RESULT:-
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 42
Review Questions:-
1) The internal resistance of a source is 2 Ohms and is connected with an
External load of 10 Ohms resistance. What is Rth ?
2) In the above question if the voltage is 10 volts and the load is of 50Ω.
What is the load current and Vth? Verify IL?
3) If the internal resistance of a source is 5 Ω and is connected with an
External load of 25 Ohms resistance. What is Rth?
4) In the above question if the voltage is 20V and the load is of 50 Ohms,
What is the load current and IN ? Verify IL ?
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 43
OPEN CIRCUIT CHARACTERISTICS OF D.C SHUNT GENERATOR
AIM:
To obtain the no load characteristics of a DC shunt generator and to determine the critical
field resistance.
NAME PLATE DETAILS:
S.NO Characteristic D.C Motor D.C Generator
1 Voltage 220V 220V
2 Current 19. A 13.6 A
3 Speed 1500 R.P.M 1500 R.P.M
4 Power 5 HP 3 KW
5 Field current 1 A 1 A
APPARATUS REQUIRED:
S.NO Description Type Range Quantity
1 Volt meter M.C 0-300v 1
2 Ammeter M.C 0-2A 1
3 Tachometer Digital 0-10,000 R.P.M 1
4 Rheostat Wire wound 300Ω/2A
2
Exp - 6
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 44
5 Connecting
wires ------ ------- As required
CIRCUIT DIAGRAM:
OPEN CIRCUIT CHARACTERISTICS OF D.C SHUNT GENERATOR
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 45
THEORY:
Magnetization curve is relation between the magnetizing forces and the
flux density B. this is also expressed as a relation between the field current and the induced
emf , in a D.C machine. Varying the field current and noting corresponding values of
induced emf can determine this.
For a self-excited machine the theoretical shape of the magnetization
curve is as shown in the figure. The induced emf corresponding to residual magnetism
exists when the field current is zero. Hence the curve starts, a little above the origin on y-
axis. The field resistance line Rsh is a straight-line passing through the origin.
PROCEDURE:
1) All the connections are as per the circuit diagram.
2) 220V, DC supply is given to the motor by closing DPST switch.
3) Move the 3-point starter handle form ‘OFF’ to ‘ON’ position slowly and motor starts
running.
4) Adjust the speed of the motor to rated value by the adjusting the field rheostat of
motor.
5) By using field rheostat vary the field current of generator.
6) By varying the filed current in steps note down all the readings of generated voltages
at constant speed.
7) Now the field current & field rheostat of motor is removed slowly and the power is
switched OFF.
TO FIND CRITICAL FIELD RESISTANCE:
1) Draw the shunt field resistance line
2) Draw tangent to the OCC
3) The slope of this tangent gives the Rfc
Critical field resistance, Rc=Eg/ IF =
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 46
TABULAR COLUMN:
Residual Voltage = Speed=
GRAPH:
Draw the graph between generated voltage at no load and field current. By taking
generated voltage Eg in volts on Y axis and field current If in amps on X-axis.
S.NO Eg in Volts If in Amps
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 47
MODEL GRAPH:
TO FIND FIELD RESISTANCE:
S.NO Voltage Field current Field resistance
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 48
PRECAUTIONS:-
1) The rheostat is connected such that minimum resistance is included in field circuit of
motor
2) The rheostat is connected such that maximum resistance is included in field circuit of
generator.
3)Starter handle is moved slow
RESULT:
.
REVIEW QUESTIONS:
1.What is meant by critical field resistance?
2.What is meant by critical speed?
3.Residual magnetism is necessary for self excited generators or not.
4.Why this test is conducted at constant speed?
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 49
SWINBURNE’S TEST ON D.C SHUNT MACHINE
AIM:
To conduct the field test on two identical series machines and to find the efficiency at full
load.
NAME PLATE DETAILS:
S.NO Characteristic D.C Motor
1 Voltage 220V
2 Current 19A
3 Speed 1500 R.P.M
4 Power 5 HP
5 Field Current 0.6 A
6 Insulation Class B
APPARATUS REQUIRED:
S.NO Description Type Range Quantity
1 Voltmeter M.C 0-300V 01
Exp – 7
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 50
2 Ammeter M.C 0-2A
0-5 A
0-1A
01
01
01
3 Tachometer Digital 0-10000 R.P.M 1
4 Rheostat Wire wound 300Ω/2A
01
5 Connecting
Wires
---- ---- As required
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 51
CIRCUIT DIAGRAM: SWINBURNS’S TEST ON D.C SHUNT MACHINE
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 52
THEORY:
It is a simple indirect method in which losses are determined separately and
from their knowledge, efficiency at any desired load can be predetermined. the only test needed
is no-load test. However, this test is applicable to those machines in which flux is practically
constant.
The constant losses in a dc shunt machine= Wc = stray losses(magnetic & mechanical
losses) +shunt field copper losses
Wc = No load input – No load armature copper losses
= VIL0 - Iao2 where Ra is the armature resistance
and Iao=IL0-Ish
PROCEDURE
1) Make all the connections are as per the circuit diagram.
2) Keep the field rheostat in minimum resistance position and armature rheostat in
maximum position.
3) Excite the motor with 220V, DC supply by closing the DPST switch and start the
motor by moving the handle of 3-point starter from OFF to ON position.
4) By adjusting the rheostats in motor armature and field bring the speed of the motor to
its rated value. Note down the readings of Ammeter and Voltmeter at no load
condition
5) The necessary calculations to find efficiency of machine as motor & generator at any
given value of armature current is done.
TO FINDARMATURE RESISTANCE(RA):
1) Connect the circuit per the circuit diagram
2) Keep the rheostat in maximum position.
3) Now excite the motor terminals by 30V supply by closing DPST switch.
By varying the rheostat & motor down the readings of Ammeter and voltmeter
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 53
MODEL CALCULATIONS:-
For motor:
IL= Ia+If
No load losses = Wo =VIo – Iao2Ra
Input = VI
Cu losses = Ia2 Ra
Total losses =No load losses + cu losses
Efficiency( η) = Output/Input
Output = input-total losses
For generator:-
I a = IL +If
No load losses = Wo = V Io – Iao2Ra
Input = VI
Cu losses = Ia2 Ra
Total losses =No load losses + cu losses
Efficiency ( η) = Output / Input
Output = input - total losses
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 54
CIRCUIT DIAGRAM TO FIND ARMATURE RESISTANCE:
TABULAR COLOUMN:
S.NO Voltmeter
reading
V Volts
Ammeter
Reading
I in Amps
Ammeter
reading
If in Amps
Speed in
RPM
ARMATURE RESISTANCE (Ra):
S.No Voltage Current
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 55
CALCULATION TABLE:
As a Motor:
S.NO Load IL in A Ia=(IL-Ish)
in A
Win =Ia2
Ra in
watts
Total
losses in
w
%Efficiency
As a Generator:
S.NO Load IL in A Ia=(IL+Ish)
in A
Win =Ia2
Ra in
watts
Total
losses in
w
%Efficiency
GRAPH:
The graph is drawn between
(a)Output in Watts Vs Efficiency(%η)
By taking output in Watts on X axis current, Efficiency on Y-axis
MODEL GRAPH:
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 56
THEORY:
Advantages:
1) It is convenient and economical method of testing of DC machines since power required
to test a large machine is very small.
2) The efficiency of the machine can be predetermined at any load. Since stray losses are
known.
Disadvantages:
1) This test cannot be performed with dc series motors.
2) This test is only applicable to those machines in which flux and speed remain
constant.
3) As the test is performed on no load it is impossible to know whether at full load
commutation would be satisfactory and the temperature raise would be with in
specified limits or not.
4) No account is taken for change in iron losses form no load to full load on account of
distribution of flux due to armature reaction. On full load the flux distribution is very
much affected due to armature reaction and is some case to an extent that iron losses
become 1.5 times of iron losses at no load .
PRECAUTIONS:
1.We should start the motor under no load
2.Take the reading without parallax error.
3.The connections must be tight.
4. If voltmeter gives –ve reading then interchange voltmeter terminal connecting of
voltmeter.
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 57
RESULT:
REVIEW QUESTIONS:
1.Why the magnetic losses calculated by this method are less than the actual value?
2.Is it applied to D.C series machines?
3.Comment on the efficiency determined by this method.
BRAKE TEST ON D.C SHUNT MOTOR
AIM:
To conduct the brake test on a given D.C shunt motor and to draw its performance
curves .
NAME PLATE DETAILS:
S.NO Characteristic D.C Motor
1 Voltage 220V
2 Current 19A
3 Speed 1500 R.P.M
4 Power 5 HP
5 Field current 0.6 A
APPARATUS REQUIRED:
Exp – 7
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 58
S.NO Description Type Range Quantity
1 Volt meter M.C 0-300v 01
2 Ammeter M.C 0-20A 01
3 Rheostat Wire wound 300Ω/2A 01
4 Tachometer Digital 0-10000 R.P.M 01
5 Connecting
Wires
---- ---- As required
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 59
CIRCUIT DIAGRAM: BRAKE TEST ON D.C SHUNT MOTOR
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 60
THEORY:
The speed of a D.C motor Nα V.Ia Ra
ϕ The speed of the D.C motor is inversely proportional to the flux produced by the
field. Decreasing the flux the speed of the machine can be increased. The flux of the field
winding can be changed by changing shunt field current with the help of shunt field rheostat.
Another method for speed control is to keep a variable resistance in series with the armature. By
increasing the resistance the voltage drop also increases and hence the voltage applied across the
armature decreases which result in the decrease in speed of the motor.
PROCEDURE:
1. All the connections are as per the circuit diagram.
2. 220V, DC supply is given to the motor by closing DPST switch.
3. Move the 3-point starter handle form ‘OFF’ to ‘ON’ position slowly and motor
starts running.
4. Vary the field rheostat and armature rheostat until the motor reaches its rated
speed and take voltmeter and ammeter readings.
5. Apply the land by break drum pulley and for each applications of load the
corresponding Voltmeter (V), Ammeter (I), Spring forces S1 & S2 and Speed (N)
readings are noted.
6. Calculate output & efficiency for each reading.
7. Note down all the readings in the tabular form carefully.
8. Remove the load slowly and keep the rheostat as starting position and switch
‘OFF’ the supply by using DPST switch
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2-1ECE ET LABMANUAL
Dept of E.E.E Page 61
Tabular column
S.NO
Voltmet
er
Reading
V volts
Amme
ter
Readi
ng I
amps
Input =
VI
watts
Forces in
KG Net forces
F = S1~S2
in kg
Torque
=
f*r*9.81
(N-M)
Speed in
RPM
(N)
O/p=
2πNT/60
(Watts)
%ή=
(o/p)/(i/p)*100 S1 S2
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 62
GRAPH:
The graph is drawn between
a) Output in Watts Vs Speed(N) in RPM
b) Output in Watts Vs Torque(T) in N-m
c) Output in Watts Vs Current(I) in A
d) Output in Watts Vs Efficiency(%η)
By taking output in Watts on X axis and speed, Torque, current, Efficiency on
Y- axis .
MODEL GRAPH:
FORMULAE:
Torque=:F*Re*9.81 N
Power output=[(2*Π*N*T) /60] W
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 63
PRECAUTIONS:
1. Initially 3-point starter should be kept at ‘OFF’ position and later it must be varied
slowly and uniformly from ‘OFF’ to ‘ON’ position.
2. The field regulator must be kept at its minimum output position.
3. The brake drum of the motor should filled with cold water.
4. The motor should be started without load.
APPLICATIONS:
• Essentially for constant speed applications requiring medium starting torque.
• May be used for adjustable speed not greater than 2:1 range.
• For lathes, centrifugal pumps, reciprocating pumps, fans, blowers, conveyors, wood working
machines, machine tools, printing presses, spinning and weaving machines etc.
RESULT:
REVIEW QUESTIONS:
1 .Why a 3-point starter is used for starting a D.C shunt motor?
2 . If a 3-point starter is not available ,how can a D.C motor be started?
3 . Explain the function of overload release coil in 3-point starter .
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 64
OC & SC TESTS ON 1 – PHASE TRANSFORMER
AIM:
To conduct Open circuit and Short circuit tests to pre-determine the performance of the
single phase transformer
NAME PLATE DETAILS:
Voltage 230 V (HV), 115 V (LV)
Current 26 A(HV) ,13.04 A (LV)
KVA RATING 3KVA
APPARATUS:
S.NO Name Type Range Quantity
1 Ammeter MI
MI
(0- 5) A,
(0- 20) A
1
1
2 Voltmeter MI
MI
(0- 300 )V,
(0- 50 )V
1
1
3 Wattmeter UPF 150 V/ 20 A 1
4 Wattmeter LPF 300 V/5 A 1
5 Auto
Transformer
1-Φ 230 V / (0- 270) V/ 15
A
1
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 65
CIRCUIT DIAGRAMS
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 66
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 67
OC Test Observations
Where
M. F. = Multiplication factor = FSD
VI φcos
FSD Full scale divisions
SC Test Observations
SVSC (V) ISC (A) WSC = W x M.F (w)
S.No. Vo (V) Io (A) Wo = W x M.F (w)
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Tirumala Engineering College Electrical Technology lab
Dept of E.E.E Page 68
PROCEDURE (OC TEST):
1. Connections are made as per the circuit diagram
2. Initially variac should be kept in its minimum position
3. Close the DPST switch
4. By varying Auto transformer bring the voltage to rated voltage
5. When the voltage in the voltmeter is equal to the rated voltage of LV winding note down all
the readings of the meters
6. After taking all the readings bring the variac to its minimum position
7. Now switch off the supply by opening the DPST switch
PROCEDURE (SC TEST):
1. Connections are made as per the circuit diagram
2. Short the LV side and connect the meters on HV side
3. Before taking the single phase, 230 V, 50 Hz supply the variac should be in minimum
position
4. Now close the DPST switch so that the supply is given
5. By varying the variac when the ammeter shows the rated current then note down all the
readings
6. Bring the variac to minimum position after taking the readings and switch off the supply
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MODEL CALCULATIONS:
(a)Calculation of Equivalent circuit parameters:
Let the transformer be the step-up transformer
Primary is L. V. side.(V1) , Secondary is H. V. side (V2)
(i) Parameters calculation from OC test
cos φφφφ0 = oo
o
IV
W =
Iw = I0 cos φφφφ0 = KII ww /1 = =
wI
VR 1
0 = = 2
0
1
0 KRR = =
Iµ = I0 sin φφφφ0 =
µI
VX 1
0 = = 2
0
1
0 KXX = =
K = 1
2
V
V =
(ii) Parameters calculation from SC test
202
sc
SC
I
WR = = 2
02
2
0202 RZX −= =
SC
SC
I
VZ =02 =
2
0201 / KXX = =
2
0201 / KRR = =
2
0201 / KZZ =
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Tabulation:
(a) Efficiency at different loads and P.f s
cos φφφφ1 = ___________ cosφφφφ2 = ___________
S.N
o.
Load Cu.lo
ss
(W)
Outp
ut
(W)
Inp
ut
(W)
η
(%
)
X
x
S.N
o.
Loa
d
Cu.lo
ss
(W)
Outp
ut
(W)
Inp
ut
(W)
η
(%
)
1.
2.
3.
4.
¼
F.
L.
½
F.
L.
¾
F.
L.
F.
L.
1.
2.
3.
4.
¼
F.L
.
½
F.L
.
¾
F.L
.
F.L
.
(b) Regulation at full load
Lagging Pf Leading Pf
S.
No. P.F. % Reg.
S.
No. P. F. % Reg.
1. 0.3 1. 0.3
2. 0.4 2. 0.4
3. 0.5 3. 0.5
4. 0.6 4. 0.6
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Dept of E.E.E Page 71
5. 0.7 5. 0.7
6. 0.8 6. 0.8
7. Unity 7. Unity
(b) Calculations to find efficiency:
For ½ full load
Cupper losses = Wsc x (1/2)2 watts =
where Wsc = full – load copper losses
Constant losses = W0 watts =
Output = ½ KVA x cos φφφφ = [cos φφφφ may be assumed]
Input = output + Cu. Loss + constant loss =
% 100xInput
Outputefficiency = =
(C)Calculation of Regulation at full load:
I2 = Load (KVA) X 103 / V2 =
100sincos
Re%2
022022 xV
XIRIgulation
φφ ±= =
‘+’ for lagging power factors
‘-‘ for leading power factors
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Dept of E.E.E Page 72
PRECAUTIONS:
1. Loose connections should be avoided
2. Variac should be in minimum position initially
3. During the SC test make sure that constant current is supplied
MODEL GRAPHS:
1) Load Vs Efficiency
2) Pf Vs Regulation
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RESULT:
REVIEW QUESTIONS:
1) The regulation calculated is exact or approximate?
2) Is it direct or indirect test?
3) What are the parameters to be calculated by using this test?
4) What are the conditions for maximum regulation and zero regulation?
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BRAKE TEST ON 3 -PHASE SQUIRREL CAGE INDUCTION MOTOR
AIM:
To perform Brake test on 3- phase Slip ring induction motor to determine performance
characteristics.
NAME PLATE DETAILS:
Voltage 415 V
Current 7 A
Power 3 H.P
Speed 1500 Rpm
APPARATUS:
S.NO NAME TYPE RANGE QUANTITY
1 Voltmeter MI 0-600 V 01No
2 Ammeter MI 0-10 A 01No
3 Wattmeter Dynamometer 600 V/ 10 A
UPF(F/R)
02No
4 Tachometer Digital (0-10000)RPM 01NO
5 Connecting
Wires
---- ---- Required
number
Exp – 9
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CIRCUIT DIAGRAM
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PROCEDURE:
1. Connect the circuit as per the circuit diagram
2. Insert the proper rating fuses
3. Initially we place the rotor rheostat in maximum position
4. Initially there is no load on the motor
5. Close the TPST switch by giving 415 V, 50 Hz AC supply using 3-phase Auto Transformer
6. Measure no load voltage, current and power
7. Apply the load by tightening the brake drum measure the values of voltage, current ,
wattmeter, speed, weights reading i.e. S1 and S2.
8. After taking the values make brake drum in initial position(voltage 415 to 0 position) then
switch off the supply (Remove TPST switch)
9. Tabulate the results and draw the graphs.
MODEL CALCULATION:
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MODEL GRAPHS:
1. Output Vs Efficiency
2. Output Vs Torque
3. Slip Vs Torque
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PRECAUTIONS:
1. While starting the motor the load should not be there on the drum
2. The applied load on the brake drum does not exceed its rated value
3. The readings should be noted without parallelex errors
4. Always the drum should be kept cool
RESULT:
REVIEW QUESTIONS:
1) Is it possible to realize maximum torque at starting of induction motor?
2) What should be the frequency of injected emf in rotor circuit?
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OC & SC TESTS ON 3 – PHASE ALTERNATOR
AIM:
To conduct Open circuit and Short circuit tests on a three- phase Alternator and to determine
the voltage regulation and synchronous impedance using EMF and MMF methods
NAME PLATE DETAILS:
DC SHUNT MOTOR ALTERNATOR
VOLTAGE (0-220 )V VOLTAGE: 415 V
CURRENT: 2.8 A CURRENT: 7 A
SPEED: 1500 RPM SPEED: 1500 RPM
POWER: 7.5 HP POWER: 7.5 HP
EXCITATION CURRENT: 2 A EXCITATION CURRENT: 1.55 A
APPARATUS:
S.NO NAME TYPE RANGE QUANTITY
1 Voltmeter MI 0-600 V 01No
2 Voltmeter MC 0-50 V 01No
3 Ammeter MI 0-10 A 01No
4 Ammeter MC 0-2 A 01No
5 Rheostat Wire
Wound
300 Ω/ 2 A
100Ω/ 5 A
01No
01No
6 Connecting
Wires
---- ----- Required
number
7 Tachometer
Digital
(0-
10,000)RPM
01No
Exp – 10
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CIRCUIT DIAGRAM:-
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Tabulation:
a) OC & SC Test:
O. C. Test S. C. Test
Speed = Speed =
S.No. Field
current (A)
Phase
voltage (V)
S.No. Field
current,
(If) (A)
Short circuit
current (ISC), (A)
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b) Armature Resistance:
S.No. I (A) V (volts) Rdc = V/I ΩΩΩΩ
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PROCEDURE:
(OC TEST):
1. Connect the circuit as per the circuit diagram
2. Bring the D.C shunt motor to rated speed by varying rheostat (say field).
3. Alternator is brought to the synchronous speed by using shunt motor.
4. Initially the stator terminals are open circuited.
5. Now increase (Potential divider) the field current of alternator and note the values of ammeter,
voltmeter.
6. Slowly reduce the (Potential divider ) excitation and field rheostat of shunt motor to their
initial positions. switch off supply
(SC TEST):
1. Connect the circuit as per the circuit diagram
2. The Stator terminals are short circuited using a switch
3. Now close the DPST switch and by varying the field rheostat bring DC shunt motor to its
rated speed
4. Alternator is synchronized using shunt motor
5. Now vary the DC excitation of alternator in steps to field current rated value and note A1 and
A2 values
6. Bring back the excitation and field rheostat to their initial positions
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MODEL CALCULATIONS:
EMF METHOD:
From Graph SC
OC
SI
VZ = for the same If and speed: =
Ra = (1.6) RdC =
22
aSS RZX −= =
Assume p.f. (CosΦ) =
Assume armature current (Ia) =
Generated emf of alternator on no load is
( ) ( )22
0 sincos Saaa XIvRIvE ±++= φφ =
+ for lagging p.f.
- for leading p.f.
The percentage regulation of alternator for a given p.f. is
100Re% 0 xV
VEg
−= =
where
E0 – Generated emf of alternator per phase voltage
V – Full load, rated terminal voltage per phase.
PRECAUTIONS:
1. The excitation should not exceed the field current
2. The rheostat in shunt motor should be at minimum position
3. Take the readings without parallelox errors
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MODEL GRAPHS:
1. Field current ( If ) Vs Generated EMF ( E )
2. Field current ( If ) Vs Short circuit current ( I sc )
3. Power factor Vs % Regulation
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RESULT:
REVIEW QUESTIONS:
1) What is meant infinite bus bar?
2) How to minimize hunting effect?
3) At what condition frequency collapses?
4) What is synchroscope?
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